New home wiring

JohnW2 · 13 Oct 2015

EFLImpudence said:
Yes, but the discussion is: would the circuit load be limited to 32A or could it rise to 72A?

I think it really started as a discussion as to whether it could be described as a "72A ring" - and, as I said, it's really a question of what you mean by that. Yes, as I've just written to stillp, it would be possible to arrange a 72A load such that the current through neither MCB exceeded its In, but (with other arrangements of the load) it would also be possible for a current under 40A to put more than 32A through the 32A MCB, or for a current not much more than 40A to result in more than 40A flowing through the 40A MCB.

It therefore comes down to what you want to mean by an "X amp ring circuit".

Kind Regards, John

Robin0577 · 13 Oct 2015

JohnW2 said:
EFLImpudence said:

JohnW2 said:

If one end of the ring is supplied via a 32A MCB, adding another MCB at the other end makes no difference to the distribution of currents.

Click to expand...

How does the load know from which end to draw the current?

Click to expand...

From the ratio of the impedances of the two paths back to the CU.

Consider an extreme example of a 50 metres (total length) ring, with a 64A load applied 1 metre from one end. The impedances of the two paths from load to CU will be in the ratio 49:1, hence the currents through those two paths will also be in the ratio 1:49. With 64A load, that means that about 62.72A will flow through the nearest MCB and about 1.28A through the one at t'other end - totally regardless of what OPDs might be fitted to the two ends.

Kind Regards, John

Hang on a minute, aren't you assuming a zero impedance source for that calculation? Is the current sharing not dependant on the total impedance of the circuit? The higher the source impedance gets, the less difference the impedance of the "ring" makes. Say you have an external source impedance of 0.3 ohm and a cable impedance of 0.75 ohm/metre, then you have a total impedance of 0.3075 ohm on the 1m leg and 0.435 on the 49m leg, a ratio of 41:58.

JohnW2 · 14 Oct 2015

Robin0577 said:
Hang on a minute, aren't you assuming a zero impedance source for that calculation? Is the current sharing not dependant on the total impedance of the circuit? The higher the source impedance gets, the less difference the impedance of the "ring" makes. Say you have an external source impedance of 0.3 ohm and a cable impedance of 0.75 ohm/metre, then you have a total impedance of 0.3075 ohm on the 1m leg and 0.435 on the 49m leg, a ratio of 41:58.

Firstly, I can't make a lot of sense of your example figures.

... but, no, you are overlooking the fact that the two legs are joined both at at the CU(s) and at the load (i.e. are simply in parallel), so the potential difference across the impedances of each of the legs of the ring must be the same. If the impedance of the short leg of the ring were, say, 0.01Ω and the impedance of the long leg hence 0.49Ω, if we call the common pd across both impedances Vx volts, then the currents through the two impedances/legs would be Vx/0.01 and Vx/0.49 amps respectively, a ratio of 49:1 - regardless of what was happening upstream of the CU(s) (common to both legs of the ring).

Kind Regards, John

Robin0577 · 14 Oct 2015

Yes of course, i knew I was overlooking something, and the junction at the CU was it. I had separate source paths in my head for some stupid reason. I'll get my coat

stillp · 14 Oct 2015

Meanwhile, back in the real world...

JohnW2 · 14 Oct 2015

stillp said:
Meanwhile, back in the real world...

Quite so - this was a particularly outrageous tangent! However, we can but believe that the case Adrian cited was from the real world, although I don't know how certain he is that the motivation/intent of the perpetrator was necessarily quite as he suggested!

However, I can but presume that it has been somewhat educational for some, since there clearly were some who, at least initially, believed that adding a second, higher-rated, MCB at the second end of a ring final would alter the current/cable situation.

Kind Regards, John

EFLImpudence · 14 Oct 2015

JohnW2 said:
since there clearly were some who, at least initially, believed that adding a second, higher-rated, MCB at the second end of a ring final would alter the current/cable situation.

Did you leave out a "not" ?

JohnW2 · 14 Oct 2015

EFLImpudence said:
JohnW2 said:

since there clearly were some who, at least initially, believed that adding a second, higher-rated, MCB at the second end of a ring final would alter the current/cable situation.

Click to expand...

Did you leave out a "not" ?

No

Kind Regards, John

EFLImpudence · 15 Oct 2015

Unusually curt for you.

Please explain why a 'ring' on two separate MCBs does not alter the current/cable situation.

JohnW2 · 15 Oct 2015

EFLImpudence said:
Unusually curt for you.

It seemed adequate!

EFLImpudence said:
Please explain why a 'ring' on two separate MCBs does not alter the current/cable situation.

For a given set of loads plugged into a ring final circuit, the current in every part of every cable will be identical regardless of what, if any, OPDs are connected to one, both, or no ends of the ring (unless, of course, one of the OPDs operates!).

That's pretty curt, too

Kind Regards, John

EFLImpudence · 15 Oct 2015

Nope, I must be missing something.

You are surely not saying it is safe, are you?

JohnW2 · 15 Oct 2015

EFLImpudence said:
Nope, I must be missing something. You are surely not saying it is safe, are you?

Of course I'm not. I was merely pointing out that, contrary to what appeared to be being implied, the presence/absence/rating of one or two OPDs protecting a ring does not, per se, have any effect on currents flowing as a result of a particular loading.

Provided the total load did not exceed 32A, it would obviously be no less safe (other than possibly during fault conditions) than a standard ring with the same loads applied. However, the splitting of the current between two RCDs (even two 20A ones) obviously opens up the possibility of a sufficiently large load to be applied (without any OPD operating) so as to appreciably overload some, or all of the cable. The need for 'dual isolation' to make the circuit dead would be an added 'unsafe' aspect of such an arrangement.

Kind Regards, John

EFLImpudence · 15 Oct 2015

Right, so it does alter the current/cable situation.

It could allow 58A (40x1.45) for an hour etc. on part of what could be a 20A rated cable.

JohnW2 · 15 Oct 2015

EFLImpudence said:
Right, so it does alter the current/cable situation.

Ah - language issues again! It does not alter the current in the cables!

EFLImpudence said:
It could allow 58A (40x1.45) for an hour etc. on part of what could be a 20A rated cable.

Indeed so - although, to be fair, that "20A rated cable" is deemed to be able to safely carry at least 29A (20A x 1.45) for an hour, so it's not quite as bad as you imply!

Of course, even with a conventional ring, if the load is all very close to one end of the ring, then the 32A MCB could allow nearly 46.4A (32A x 1.45) to flow through that short length of cable (which, again, could be "20A rated cable") for an hour.

Any situation in which the In of the OPD exceeds the CCC of the cable obviously carries the risk of some degree of cable overload, but it has been deemed that a 32A OPD protecting a 20A cable in a ring final is usually "OK". Any increase in the OPD's In beyond that obviously increases the risk and, even forgetting the regs, I don't think that many people would be very happy with an In greater than 32A (for a potentially 20A cable).

Kind Regards, John

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